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I was recently solving problems related to Alcohol.

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and its solution is

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My doubt is why methanol act as a nucleophile here and when does alcohol acts as a nucleophile and as a Solvent?

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    $\begingroup$ On an unrelated note, the last step of the synthesis makes so little sense. You are using lithium aluminium hydride with water? LAH is pyrophoric and should never be used with water. Secondly, the production of an acid upon reduction of an ester is equally suspicious. $\endgroup$ – Tan Yong Boon Jan 3 at 1:42
  • $\begingroup$ @TanYongBoon Actually It is a question from a book of Advanced Problems in Organic Chemistry by Himanshu Pandey for the competitive exam "JEE". I think they add LAH after removing Ester from the water but yes I didn't observe the reduction of the ester. I don't know why the author has proceeded reaction like that. I need to ask another question about it. $\endgroup$ – CHEMOJEE Jan 3 at 3:39
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    $\begingroup$ C5H8O3 (as does C6H10O3) has 2 degrees of unsaturation -- namely -- levulinic acid (4-oxo-pentanoic acid). The final product from LiAlH4 reduction is the starting diol, both labeled "A". The solution presented is incorrect. $\endgroup$ – user55119 Jan 3 at 15:21
  • $\begingroup$ @user55119 Yes. I think it should be a printing mistake as the reaction with LAH will make the ester part to alcoholic part so that therefore we would get a diol i.e A $\endgroup$ – CHEMOJEE Jan 4 at 5:27
  • $\begingroup$ My main concern is that the second structure in the "solution" is not a hydroxy acid but the keto acid, levulinic acid. $\endgroup$ – user55119 Jan 4 at 14:59
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Welcome on Chemistry SE!

Here, you are doing an esterification, done with your carboxylic acid and methanol with $\ce{H+}$. The problem is, the yield for such reaction is quite low (an average 67% for a primary alcohol, because of the equilibrium between esterification and hydrolysis). A solution to increase it is to use an excess of reagant (preferably the cheapest one). And the simplest way to have an excess of methanol (that is really easy and cheap to get) is to use it as a solvent.

Hope this helps! And for further explanations, see Fischer-Speier esterification.

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    $\begingroup$ You have clarified the first part but the main thing I need is the general case. $\endgroup$ – CHEMOJEE Jan 1 at 14:48
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    $\begingroup$ As we’re talking about organic chemistry, it is quite hard to get a general case, it really depends on the reaction and the context... nevertheless, in general when the alcohol is the only molecule mentioned in a step of a reaction you’d better verify if your reagents can react with your alcohol (most of the time it will be the case). In any other cases, simply keep in mind that a common convention is to write the reagents above the reaction arrow and the solvent below it. $\endgroup$ – Thomas Prévost Jan 1 at 17:06
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The proposed solution to this roadmap problem has several errors. I am assuming that $\ce{K2Cr2O7}$, methanol and $\ce{LiAlH4}$ are employed in excess. In the oxidation of 1 (A), chromic acid, $\ce{H2Cr2O7}$, is formed from $\ce{K2Cr2O7}$. To obtain the purported hydroxy acid 6 suggested in the solution, exclusive oxidation of the primary alcohol would be required, a pathway that leads to lactone 5 via hemiacetal 4. One could argue that hydrolysis of the lactone occurs to form 6 but, if this were the case, then hydroxy acid 6 would oxidize to levulinic acid 2 in the presence of excess oxidant. Clearly, levulinic acid 2 fits the formula $\ce{C4H8O3}$ presented in the original problem. As the problem is set forth, oxidation of the secondary alcohol is faster than that of the primary alcohol of diol 1.

The Fischer esterification of levulinic acid 2 leads to methyl levulinate 3 ($\ce{C6H10O3}$). Such an esterification is invariably conducted with excess methanol. The reduction of 3 with $\ce{LiAlH4}$ is conducted in a solvent such as ether or tetrahydrofuran (step 1) and then decomposed cautiously with water (step 2) to reform diol 1 (A) by reduction of both the ketone and ester groups of 3. Clearly, this sequence of transformations is an academic exercise and not one to be conducted in a laboratory. Hopefully, the person who provided the solution was not the same person as the one who created the question.

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  • $\begingroup$ Thank you very much. $\endgroup$ – CHEMOJEE Jan 5 at 5:56

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